Heat Transfer Through a Plane Wall

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Homework Help Overview

The discussion revolves around a heat transfer problem involving a large plane wall with specified dimensions and thermal properties. The original poster is tasked with evaluating the rate of heat transfer through the wall, given certain boundary conditions and heat transfer coefficients.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to solve the problem using differential equations and boundary conditions, while others explore the concept of thermal resistance to arrive at similar results. There is a question regarding the correctness of the textbook's answer and the appropriateness of temperature units used in the calculations.

Discussion Status

Participants have shared their approaches and findings, noting that they arrived at the same heat transfer rate. There is an ongoing exploration of potential discrepancies with the textbook answer, and some participants express a willingness to verify assumptions and clarify concepts as they learn.

Contextual Notes

There is a mention of the textbook being used for the course, and a participant questions the necessity of using Kelvin for temperature, which leads to a discussion about the interchangeability of Celsius and Kelvin in the context of temperature differences.

bob1182006
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Homework Statement


Consider a large plane wall of thickness L = 0.4m, thermal conductivity k = 1.8W/(m*K), and surface area A = 30m2. The left side of the wall is maintained at a constant temperature of T1 = 90 C while the right side looses heat by convection to the surrounding air at Ts = 25 C with a heat transfer coefficient of h = 24 W(m2*K). Assuming constant thermal conductivity and no heat generation in the wall evaluate the rate of heat transfer through the wall.
Answer: 7079 W

My answer doesn't match the book's answer.

Homework Equations


\dot{Q}_{wall} = -kA\frac{dT(0)}{dx}
\frac{d^2T}{dx^2} = 0
T(0) = 90
-k\frac{dT(L)}{dx} = h[T(L) - Ts]

The Attempt at a Solution



Solving the differential equation and applying B.C.:
T(x) = xC_1 + C_2
T(0) = C_2 = 90
-kC_1 = hLC_1+hC_2-25h]
C_1 = -\frac{h(C_2-25)}{k+hL}

Plugging numbers in:
C_2 = 90
C_1 = -136.8
T(x) = 90-136.8x
\dot{Q}_{wall} = -1.8*30*(-136.8) = 7,387 W

Did I make a mistake or is the book's answer wrong?
 
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I used a somewhat different approach (underlying equations are of course the same) and got the same answer you did.

I computed the thermal resitance of the wall as L/kA = 7.41e-3 K/W and the equivalent thermal resistance of the convection effect as 1/hA = 1.39e-3 K/W, giving total thermal resistance of 8.80e-3 K/W so dQ/dt = (90 - 25)/8.80e-3 = 7387W.

Just curious - what is your textbook?
 
Awesome thanks, I didn't want to waste any more time redoing this problem looking for a mistake, and my class is just starting the thermal resistance chapter so if I see more problems that have wrong answers I'll do them that way to check.

We're using Heat and Mass Transfer Fundamentals & Applications 4th Ed by Cengel and Ghajar.
 
arent we suppose to take temp. as kelvin here?
 
spree said:
arent we suppose to take temp. as kelvin here?

It's OK to interchange K and C as long as you're dealing with temperature differences.
 
oh ure right. new to the subject, just trying to learn sorry :) btw I personally know professor cengel and i will inform him about this. I will let you know about it. if you suspect any other mistake in the book feel free to ask please.
 

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